Someone please help ASAP. I would really really appreciate it if u can. I will mark brainliest

What transformation was not done to the linear parent function, f(x) = x, to

Butoxors [25]

Answer:

C.

Step-by-step explanation:

The function shifted <em>left</em> five units instead of <em>right</em> five units.

There's no vertical compression in the equation provided, but that's probably just a typo since there's a random bracket that I assume was supposed to be a fraction.

7 0

2 years ago

The probability distribution for a

Tpy6a [65]

Answer:

$P(x \le -3) =0.30$

Step-by-step explanation:

Given

The attached probability distribution of x

Required

Find $P(x \le -3)$

To do this, we consider all values of x less than or equal to-3.

So, we have:

$P(x \le -3) =P(x=-5) + P(x = -3)$

From the table:

$P(x=-5) = 0.17$

$P(x = -3) = 0.13$

So, the equation becomes

$P(x \le -3) =P(x=-5) + P(x = -3)$

$P(x \le -3) =0.17 + 0.13$

$P(x \le -3) =0.30$

3 0

3 years ago

An archer is able to hit the bull's-eye 53% of the time. If she shoots 10 arrows, what is the probability that she gets exactly

Vesnalui [34]

Answer:

P(x=4)=.2458

hope it helps!

6 0

3 years ago

I have a friend who is really stressed and depressed and couldn't get this done. it's 2am for them so they're going to sleep. I

lisabon 2012 [21]

Answer:

Step-by-step explanation:

first problem:

3 is a solution to x^2=9, so to a, b, c and d

second problem

-2 is a solution to:

a, b, c and d because (-2)^2=4, 4=4, b) 5*(-2)^2=5*4=20 and so on

basically to all choices

the third, solve the equations:

x^2=1, $\sqrt{x^2}$ = $\sqrt{1}$ , x=1

x^2=81 sqrtx^2=sqrt81, x=+and -9

4x^2=1 divide by 4 , x^2=1/4, x=+&- 1/2 (plus si minus 1/2, amindoua solutiile sint valabile)

3x^2=75, we divide by 3, we get x^2=25, x=+&-5

49x^2=121 , x=+and- 11/7

25x^2=225 divide by 25, x^2=9, x =+and - 3

36x^2=144 we divide by 36, we get x^2=4, x = +and- 2

x^2-0.25=0 x^2=0.25, x=+and - 1/2

2x^2-0.32=0 , 2x^2=0.32, x^2=0.16 ( i divided by 2) , x=+and- 2/5

4x^2=0.04, divide by 4, x^2=0.01, x=+and - 0.1

3x^2=10.8, divide by 3, x^2=3.6, x=+and - 1,89

for the : determinati solutiile ecuatiilor

a) (x-1)^2=49, do the quadratic, you get x=8 and -6

due to the length of this page, i cant explain every step

b) (2x+1)^2=81, you get x= 4 and -5

c)(3x-1)^2=121, do the quadratic, x=- 10/3 and +4

d) 3(x-3)^2=48, we divide by 3 first , (x-3)^2=16, we get x=7 and -1

e)5(x+2)^2=45 we divide by 5, (x+2)^2=9, x=+1and -5

f) 7(2x-1)^2=175 we divide by 7 and we have (2x-1)^2=25 x=+3 and -2

Determine x belong to R

a) 1/9 x^2=16, we multiply 9*16=144, x^2=144, x=+and - 12

1/100x^2-1/64=0, or 1/100 x^2=1/64, x=+and - 5/4

1/25 x^2=0.49, x^2=0.49*25=12.25, x=+ and - 7/2

8 0

3 years ago

Suppose parts (a) through (d) below provide results for a study on the role of calcium in reducing the symptoms of PMS. For each

frutty [35]

Answer:

Step-by-step explanation:

Hello!

To test if calcium reduces the symptoms of PMS two independent groups of individuals are compared, the first group, control, is treated with the placebo, and the second group is treated with calcium.

The parameter to be estimated is the difference between the mean symptom scores of the placebo and calcium groups, symbolically: μ₁ - μ₂

There is no information about the distribution of both populations X₁~? and X₂~? but since both samples are big enough, n₁= 228 and n₂= 212, you can apply the central limit theorem and approximate the sampling distribution to normal X[bar]₁≈N(μ₁;δ₁²/n) and X[bar]₂≈N(μ₂;δ₂²/n)

The formula for the CI is:

[(X[bar]₁-X[bar]₂) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} }$ ]